How do you solve the trigonometric equation 2tanx = 1/2tan2x?

[JohanX]

Homework Statement

Solve for X

Relevant Equations

2tanx= 1/2 tan2x

2tanx= 1/2tan2x

4tanx=2tanx / 1 - tan^2x

1- tan^2x = 2tanx / 4tanx

tan^2x = 1/2

tanx = √1/2

Textbook answer 0 ; +/- 0.615 rad

Not sure where I am going wrong.

 

[PeroK]

Homework Statement: Solve for X
Homework Equations: 2tanx= 1/2 tan2x

2tanx= 1/2tan2x

4tanx=2tanx / 1 - tan^2x

1- tan^2x = 2tanx / 4tanx

tan^2x = 1/2

tanx = √1/2

Textbook answer 0 ; +/- 0.615 rad

Not sure where I am going wrong.

What trig identities are you using?

 

[JohanX]

Im using the double angles, half angles:

sin^2x + cos ^2x = 1
sin(a+/-b)= sinAcosB +/- cosAsinB
cos(a+/b)= cosAcosB -/+ sinAsinB
tan(a+/-b)= tanA+/-tanB / 1-/+ tanAtanB

sin2A = 2sinAcosA
cos2A= cos^2A - sin^2A
cos2A= 1 - 2sin^2A
tan2A= 2tanA / 1 - tan^2A

sin A/2 = √1/(1-cosA)
cosA/2= √1/(1+cosA)
tanA/2= √1-cosA/1+cosA

In this question I used tan2A identity.

 

[Delta2]

Well, you have solve it and you haven't realized it.

$tanx=\frac{\sqrt{2}}{2}\Rightarrow x=tan^{-1}\frac{\sqrt{2}}{2}=0.615rad$

The only step you omit is that when you divide by $tanx$ you also have to take the case where
$tanx=0\Rightarrow x=0$

 

[Mark44]

Homework Statement: Solve for X
Homework Equations: 2tanx= 1/2 tan2x

2tanx= 1/2tan2x
4tanx=2tanx / 1 - tan^2x
<snip>

Please use parentheses. In the first equation, 1/2tan2x means $\frac 1 2 \tan(2x)$. If you don't use LaTeX, it should be written as 1/(2 tan(2x)).
In the second equation, what you wrote on the right side means $\frac{2\tan(x)}1 - \tan^2(x)$, which is probably not what you meant. As inline text, this should be 2 tan(x)/(1 - tan^2(x))

In the lower left corner there's a link to our LaTeX tutorial.

 

[JohanX]

Thanks for the help